For $g\isin\Z$ , let $\bar{g}\isin\Z_{37}$ denote the residue class of g modulo 37. Consider the group U 37 = { $\bar{g}\isin\Z_{37}:1\le g\le37$ with gcd(g, 37) = 1} with respect to multiplication modulo 37. Then which one of the following is FALSE?

The set ${\bar g \isin U_{37}: \bar g = (\bar g)^{-1}}$ contains exactly 2 elements.

The order of the element $\overline{10}$ in U 37 is 36.

There is exactly one group homomorphism from U 37 to ( $\Z$ , +).

There is exactly one group homomorphism from U 37 to ( $\mathbb{Q}$ ,+).

The order of the element $\overline{10}$ in U 37 is 36.

The set ${\bar g \isin U_{37}: \bar g = (\bar g)^{-1}}$ contains exactly 2 elements.

The order of the element $\overline{10}$ in U 37 is 36.

There is exactly one group homomorphism from U 37 to ( $\Z$ , +).

There is exactly one group homomorphism from U 37 to ( $\mathbb{Q}$ ,+).

Question:

For $g\isin\Z$ , let $\bar{g}\isin\Z_{37}$ denote the residue class of g modulo 37. Consider the group U₃₇ = { $\bar{g}\isin\Z_{37}:1\le g\le37$ with gcd(g, 37) = 1} with respect to multiplication modulo 37. Then which one of the following is FALSE?

Updated On: Oct 1, 2024

The set ${\bar g \isin U_{37}: \bar g = (\bar g)^{-1}}$ contains exactly 2 elements.
The order of the element $\overline{10}$ in U₃₇ is 36.
There is exactly one group homomorphism from U₃₇ to ( $\Z$ , +).
There is exactly one group homomorphism from U₃₇ to ( $\mathbb{Q}$ ,+).

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The correct option is (B): The order of the element

\overline{10}

in U₃₇ is 36.

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